Suppose I have a list that I wish not to return but to yield values from. What is the most pythonic way to do that?
Here is what I mean. Thanks to some non-lazy computation I have computed the list ['a', 'b', 'c', 'd'], but my code through the project uses lazy computation, so I'd like to yield values from my function instead of returning the whole list.
I currently wrote it as following:
my_list = ['a', 'b', 'c', 'd']
for item in my_list:
yield item
But this doesn't feel pythonic to me.
4 Answers
Since this question doesn't specify; I'll provide an answer that applies in Python >= 3.3
If you need only to return that list, do as Anurag suggests, but if for some reason the function in question really needs to be a generator, you can delegate to another generator; suppose you want to suffix the result list, but only if the list is first exhausted.
def foo():
list_ = ['a', 'b', 'c', 'd']
yield from list_
if something:
yield this
yield that
yield something_else
In versions of python prior to 3.3, though, you cannot use this syntax; you'll have to use the code as in the question, with a for loop and single yield statement in the body.
Alternatively; you can wrap the generators in a regular function and return the chained result: This also has the advantage of working in python 2 and 3
from itertools import chain
def foo():
list_ = ['a', 'b', 'c', 'd']
def _foo_suffix():
if something:
yield this
yield that
yield something_else
return chain(list_, _foo_suffix())
Use iter to create a list iterator e.g.
return iter(List)
though if you already have a list, you can just return that, which will be more efficient.
You can build a generator by saying
(x for x in List)
Best Pythonically way to do this task is
In [1]: my_list = ['a', 'b', 'c', 'd']
In [2]: (x for x in my_list)
Out[2]: <generator object <genexpr> at 0x7fd9e4fdc230>
